Introduction to Inequalities Roman Kvasov
Idea #1 :
Perform equivalent transformations of the initial inequality to bring it to the inequality that is obvious or known to be true
Problem Prove the inequality: a 2 + b 2 ≥ 2ab
Solution Start with the initial inequality and perform equivalent transformations passing all terms to one side and completing the square of the obtained expression: 2 a 2 + b 2 ≥ 2ab ⇔ a 2 + b 2 − 2ab ≥ 0 ⇔ ( a − b )( a − b ) ≥ 0 ⇔ ( a − b ) ≥ 0 Since the last inequality is obviously true then the initial inequality is true as well. 2 Note that the equality is reached only when ( a − b ) = 0 or equivalently when a = b .
Practice Problems 1. a 2 + 9b 2 ≥ 6ab , for any a, b 2. 2a 2 + b 2 + c 2 ≥ 2a ( b + c ) , for any a, b, c 3. a 2 + b 2 ≥ 2 ( a + b − 1) , for any a, b
4. a 2 + 6ab + 10b 2 ≥ 0 , for any a, b 5. a + b 2
2
( a + b) ≥
2
, for any a, b 2 1 1 6. a 2 + 2 ≥ a + , for any a a a
7. ( a 2 − b 2 ) ≥ 4ab ( a − b ) , for any a, b 2
2
8. ( a 3 − b3 ) ( a − b ) ≥ 3ab ( a − b ) , for any a, b 2
9. a 4 + b 4 ≥ a 3b + b3a , for any a, b
10. ( a 2 + b 2 )( c 2 + d 2 ) ≥ ( ac + bd ) , for any a, b, c, d 2
11.
2
≤ ab ≤
a+b a 2 + b2 ≤ , for a, b > 0 2 2
1 1 + a b 12. ( a + c )( b + d ) ≥ ab + cd , for any a, b, c, d > 0